Example 5.2 Astronomical measurements show that the orbital speed of masses in spiral galaxies rotating about their centers is approximately constant as a function of distance R from the galaxy center, as in the figure (for the Andromeda Galaxy). Show that this is inconsistent with the galaxy having its mass concentrated at its center & can be explained if the galaxy mass increases with distance R from the center.

Solution: Assume that the galaxy mass M has a spherically symmetric distribution ρ. We can solve the problem, despite the fact that the radius R might be hundreds of light years! The orbital speed v of a small mass m in orbit at R about the galaxy center is found (Phys. I!) by setting the gravitational force = (mass)  (centripetal acceleration) or: [(GMm)/)(R2)] = [m(v2)/R]  v = (GM/R)½ (1) or: v R-½ (2) If (2) is true, v vs. R would be like the dashed curve! For the solid (expt.) curve to be consistent with (1) requires M = M(R)  R  CONCLUSION: There must be more matter in a galaxy than is observed (“dark matter” = ~90% of mass in the universe). At the forefront of research.

Example 5.3 Consider a thin, uniform circular ring of radius a & mass M. A small mass m is placed in the plane of the ring. Find a position of equilibrium & determine if it is stable.

Use the result for a Line Distribution: (M = ∫ρ(r)ds, Φ = - G∫[ρ(r)ds/r] Here, ρ = const. = ρ  M = 2πaρ The potential at the position of a point mass m a distance r from the ring center & a distance b from the differential mass dM: Φ(r) = -ρaG∫(dφ/b) φ = angle shown in the figure, limits: (0  φ  2π) From the figure, b2 = a2 + (r)2 – 2arcosφ  Φ = - ρaG∫dφ[a2 + (r)2 – 2arcosφ]-½ = - ρG∫dφ[1 + (r/a)2 – 2(r/a)cosφ]-½

Φ = - ρG∫dφ[1 + (r/a)2 – 2(r/a)cosφ]-½ (1) Consider positions close to the center point (near r = 0): Expand (1) for r << a & integrate term by term: Φ(r)  -ρG∫dφ[1 + (r/a)cosφ + (r/a)2(3cos2φ -1)+….] (For those who know & care, the integrand in this form is a series of Legendre Polynomials!) Φ(r)  -(MG/a)[1 + (r/a)2 +…] The potential energy of mass m at r is: U(r) = mΦ(r)  -m(MG/a)[1 + (r/a)2 +…]

U(r) = mΦ(r)  -m(MG/a)[1 + (r/a)2 +…] The equilibrium position is given by (dU/dr) = 0 = -m(MGr)/(2a3)  r = 0 is an equilibrium position! This should be obvious by symmetry! Stability (or not) is obtained from the 2nd derivative: (d2U/dr2)0 = -m(MG)/(2a3) < 0  r = 0 is a position of unstable equilibrium! (“Not obvious”!)